Two identical balls (mass m) undergo a collision. Initially one ball is stationary, the other has kinetic energy of 8J. The collision is partially inelastic with 2 J energy converted to heat. What is the maximum deflection angle (alpha or beta) at which one of the balls is observed?

I have come up with the following relationships using conservation of momentum and energy:

1. Vo = V1 cos a + V2 cos B

2. 0 = V1 sin a - V2 sin B

and 12 = m (V1f^2 + V2f^2) which i believe can be written

3. 0.75Vo = V1^2+V2^2

by rearranging initial kinetic energy of ball one to solve for m.

How do i go about comparing alpha and beta? Are my equations correct
Help please! thanks

now i do not know how to compare [itex]\alpha[/itex] and [itex]\beta[/itex]

any help

It has to have [itex]\beta[/itex] in terms of [itex]\alpha[/itex] ONLY. You can do this with these equations: determine v2f in terms of v1f using 2. and substitute into 1. to find v1f in terms of [itex]\alpha \text{ and } \beta[/itex]. Substitute also into 3. to find v1f. [itex](v_0^2 = 16/m)[/itex]. Then combine the two to find [itex]beta[/itex] in terms of [itex]\alpha[/itex] (the m falls out).

I really want to get this one done but it just isnt working out
Thanks for all your help on this one

I told you it was non-trivial. Have a look at the solution for an elastic collision at: http://rustam.uwp.edu/201/L12/lec12_w.html (scroll down to the collision in 2 dimensions). For elastic collisions the two angles add to 90 degrees. The question is asking how the sum of the two angles in an inelastic collision where 25% of the energy is lost compares to 90 degrees.